Jul 22, 2019 · The assignment nums = nums[::-1] would be equivalent to create a new list in memory (with the elements reversed), and changing the pointer nums to point to that new list.

Jul 22, 2022 · A series of expressions connected by or s evaluates to the leftmost expression that has a "truthy" value - which for numeric types, means "nonzero". So this is equal to nums[I - 1] if that isn't …

Apr 12, 2025 · nums[:] is slice notation that gives a list [nums[0], nums[1], ... , nums[-1]], i.e. a copy of the list. The important distinction between nums[:] = unique and nums = unique is that the latter will …

May 8, 2019 · That last row is an obscure way of writing an if then else expression. [0, size-1] creates a list of two elements. nums[0] < nums[size-1] returns either True or False when used as a list index, …

Jun 17, 2019 · " nums : List[int] " states that nums is the name of the list function parameter/variable of type int " target: int " is another function parameter/variable of type int.

I was working on TwoSum problem of LeetCode Given an array of integers nums and an integer target, return indices of the two numbers such that they add up to target.

Whereas in first case you are copying list nums to vals and both vals and nums are pointing out to same heap storage location. So changes made to vals will automatically affects to nums.

Jul 25, 2020 · In nums[i ++] = nums[i + 1] If the i++ happens before the nums[i + 1], i + 1 will be out of bounds on the last iteration. In that case, Java rightly causes an error, and C, because it doesn't do …

Apr 5, 2019 · It is a single element, because num.lengh - 1 is an int and just gives you the last accessible index of the array. If you check wether the length of the array is > 0, then you can safely …

Nov 19, 2015 · What does int nums[5]; do? When I pass nums to std::cout, it prints a memory address I think, but I don't understand what the code itself is actually doing when it runs.